Exam COPY Flashcards

1
Q

t is free for x in φ

A

if no free x leaf in φ occurs in the scope of ∀y or ∃y for
any variable y occurring in t.

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2
Q

Draw parse tree of following formula and give which variabels are bound or free

∀x ((P (x) → Q(x)) ∧ S(x, y))

A
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3
Q

What is the meaning of following notation

φ[t/x]

A

replacing all free occurrences of x in φ
by t

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4
Q
A
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5
Q

2017 - 1b

A
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6
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7
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8
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9
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10
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11
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12
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13
Q
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14
Q

Free and bound variable

∀x ((P (x) → Q(x)) ∧ S(x, y))

A
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15
Q

establishing the validity of the sequent

Premiss x + 1 = 1 + x

Premiss (x + 1 > 1) → (x + 1 > 0)

Conclusion (1 + x) > 1 → (1 + x) > 0

A
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16
Q

Premiss ∀x (P (x) → Q(x))

Premiss ∃x P (x)

Conc. ∃x Q(x)

A
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17
Q

Prove folowing sequent

Premiss ∀x (P (x) → Q(x))

Premiss ∀x P (x)

Conc. ∀x Q(x)

A
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18
Q

Prove following sequent

Premiss P (t)

Premiss ∀x (P (x) → ¬Q(x))

Conc. ¬Q(t)

A
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19
Q

Prove the following sequent

Premiss ∀x .(P (x ) →Q (x ))

Conc. ∀x .¬Q (x ) →∀x .¬P (x )

A
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20
Q

Show general proof for

  1. For all introduction
  2. For all elimination
A
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21
Q

Exercise 1.2.1 - c

Premiss (p ∧ q) ∧ r

Conc p ∧ (q ∧ r)

A
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22
Q

Exercise 1.2.1 - e

Premiss q → (p → r)

Premiss ¬r

Premiss q

Conc. ¬p

A
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23
Q

Exercise 1.2.1 - f

Conc (p ∧ q) → p

A
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24
Q

Exercise 1.2.1 - h

Premiss p

Conc. (p → q) → q

A
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25
Q

Exercise 1.2.1- i

Premiss (p → r) ∧ (q → r)

Conc p ∧ q → r

A
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26
Q

Exercise 1.2.1 - j

Premiss q → r

Conc (p → q) → (p → r)

A
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27
Q

Exercise 1.2.1 - l

Premiss p → q

Premiss r → s

Conc p ∨ r → q ∨ s

A
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28
Q

Exercise 1.2.1 - n

Premiss (p ∨ (q → p)) ∧ q

Conc p

A
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29
Q

Exercise 1.2.1 - o

Premiss p → q, r → s

Conc p ∧ r → q ∧ s

A
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30
Q

Exercise 1.2.1 - r

Premiss p → q ∧ r

Conc (p → q) ∧ (p → r)

A
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31
Q

Exercise 1.2.1 - v

Premiss p ∨ (p ∧ q)

Conc p

A
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32
Q

Exercise 1.2.1 - x

Premiss p → (q ∨ r)

Premiss q → s

Premiss r → s

Conc p → s

A
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33
Q

Exercise 1.2.1 - y

Premiss (p ∧ q) ∨ (p ∧ r)

Conc p ∧ (q ∨ r)

A
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34
Q

Exercise 1.2.2 - a

Premiss ¬p → ¬q

Conc q → p

A
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35
Q

Exercise 1.2.2 - b

Premiss ¬p ∨ ¬q

Conc ¬(p ∧ q)

A
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36
Q

Exercise 1.2.2 - c

Premiss ¬p

Premiss p ∨ q

Conc q

A
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37
Q

Exercise 1.2.2 - d

Premiss p ∨ q

Premiss ¬q ∨ r

Conc p ∨ r

A
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38
Q

Exercise 1.2.2 - e

Premiss p → (q ∨ r)

Premiss ¬q

Premiss ¬r

Conc ¬p

A
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39
Q

Exercise 1.2.2 - f

Premiss ¬p ∧ ¬q

Conc ¬(p ∨ q)

A
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40
Q

Exercise 1.2.2 - g

Premiss p ∧ ¬p

Conc ¬(r → q) ∧ (r → q)

A
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41
Q

Exercise 1.2.2 - i

Premiss ¬(¬p ∨ q)

Conc p

A
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42
Q

exercise 1.2.3 - d

Conc ¬p → (p → (p → q))

A
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43
Q

Exercise 1.2.3 - n

Premiss p ∧ q

Conc ¬(¬p ∨ ¬q)

A
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44
Q
A
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45
Q
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46
Q

Prop logic - Semantics Exercise 1.4.2 - c

Compute the complete truth table of the formula

p ∨ (¬(q ∧ (r → q)))

A
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47
Q

Prop logic - semantic 1.4.4 - b

Compute the truth value on the formula’s parse tree, or specify the corresponding
line of a truth table where

p evaluates to F, q to T and the formula is

p → (¬q ∨ (q → p))

A
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48
Q

Prop logic - semantic 1.4.4 - c

Compute the truth value on the formula’s parse tree, or specify the corresponding
line of a truth table where
the formula is

¬((¬q ∧ (p → r)) ∧ (r → q))

p evaluates to F, q to T and r evaluates to T

A
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49
Q

Prop logic - semantic - 1.4.5

A formula is valid if all its valuations evaluate to true. A formula is satisfiable if it evaluates to true for at least one of its valuations.

Is the following formula valid ? Is it satisfiable ?

A
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50
Q

Prop logic - semantic 1.4.12

Premiss p → (q → r)

Conc p → (r → q)

Show that the above sequent is not valid by finding a valuation in which the truth values of the premiss formulas are true but the formula for conclusion is false

A
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51
Q

prop logic - semantic 1.4.13 a

Premiss p ∨ q

Conc p ∧ q

give examples of natural language
declarative sentences for the atoms p and q such that the premises are true,
but the conclusion false

A
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52
Q

prop logic - semantic 1.4.13 b

Premiss ¬p → ¬q

Conc ¬q → ¬p

give examples of natural language
declarative sentences for the atoms p and q such that the premises are true,
but the conclusion false

A
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53
Q
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54
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55
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56
Q
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57
Q

Prop logic - deduction 1.2.3 q

Conc (p → q) ∨ (q → r)

Using LEM

A
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58
Q
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59
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60
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61
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62
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63
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64
Q
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65
Q

Exercise 2.3.9 - a

Prove the validity of the following sequents in predicate logic, where F , G, P ,
and Q have arity 1, and S has arity 0 (a ‘propositional atom’)

∃x (S → Q(x)) |− S → ∃x Q(x)

A
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66
Q

Exercise 2.3.9 - d

Prove the validity of the following sequents in predicate logic, where F , G, P ,
and Q have arity 1, and S has arity 0 (a ‘propositional atom’)

∀x P (x) → S |− ∃x (P (x) → S)

A
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67
Q

Exercise 2.3.9 - k

Prove the validity of the following sequents in predicate logic, where F , G, P ,
and Q have arity 1, and S has arity 0 (a ‘propositional atom’)

∀x (P (x) ∧ Q(x)) |− ∀x P (x) ∧ ∀x Q(x).

A
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68
Q

Exercise 2.3.9 - l

Prove the validity of the following sequents in predicate logic, where F , G, P ,
and Q have arity 1, and S has arity 0 (a ‘propositional atom’)

Premiss ∀x P (x) ∨ ∀x Q(x)
Conc ∀x (P (x) ∨ Q(x)).

A
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69
Q

Exercise 2.3.9 - m

Prove the validity of the following sequents in predicate logic, where F , G, P ,
and Q have arity 1, and S has arity 0 (a ‘propositional atom’)

∃x (P (x) ∧ Q(x)) |− ∃x P (x) ∧ ∃x Q(x).

A
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70
Q

Exercise 2.3.9 - n

Prove the validity of the following sequents in predicate logic, where F , G, P ,
and Q have arity 1, and S has arity 0 (a ‘propositional atom’)

∃x F (x) ∨ ∃x G(x) |− ∃x (F (x) ∨ G(x))

A
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71
Q

Exercise 2.3.9 - p

Prove the validity of the following sequents in predicate logic, where F , G, P ,
and Q have arity 1, and S has arity 0 (a ‘propositional atom’)

¬∀x ¬P (x) |− ∃x P (x).

A
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72
Q

Exercise 2.3.9 - q

Prove the validity of the following sequents in predicate logic, where F , G, P ,
and Q have arity 1, and S has arity 0 (a ‘propositional atom’)

∀x ¬P (x) |− ¬∃x P (x)

A
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73
Q

Exercise 2.3.9 - r

Prove the validity of the following sequents in predicate logic, where F , G, P ,
and Q have arity 1, and S has arity 0 (a ‘propositional atom’)

¬∃x P (x) |− ∀x ¬P (x)

A
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74
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75
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76
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77
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78
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79
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80
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81
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83
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84
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85
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86
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87
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89
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102
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103
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104
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105
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106
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112
Q

Give all temporal connectives and some examples of formulas

A
  • neXt
  • Future state
  • all future states (Globally)
  • Until
  • Release
  • Weak-until
113
Q

2016 - 1a

Premiss p → q

Premiss q → r

Premiss p → s

Conc p → (r ∧ s)

114
Q

2016 - 1c

Premiss p ∨ q

Conc ¬q → p

116
Q

Show examples of LTL connectives

A

Weak until - requires that a remains true until b becomes true, but does not
require that b ever does becomes true (i.e. a remains true
forever). It follows the expansion law of until.

117
Q

2016 - 4b

Let P , S and M be unary predicates and R a binary predicate. Decide for each of the sequents below whether they are valid or not, i.e., give a
proof in natural deduction or a counter-model.

Premiss ∀x¬R(x, x)

Conc ∀x ∀y (R(x, y) → ¬R(y, x))

118
Q

2016 - 4c

Let P , S and M be unary predicates and R a binary predicate. Decide for each of the sequents below whether they are valid or not, i.e., give a
proof in natural deduction or a counter-model.

Premiss ∀x ∀y (R(x, y) → ¬R(y, x))

Conc ∀z ¬R(z, z)

119
Q

2016 - 4d

Let P , S and M be unary predicates and R a binary predicate. Decide for each of the sequents below whether they are valid or not, i.e., give a
proof in natural deduction or a counter-model.

Conc ∀x∃y R(x, y) ∨ ∀x∃y ¬R(x, y)

123
Q

2016- 5b

Let P, Q, and R be unary predicate symbols, and f a unary function
symbol. Give proofs in natural deduction of the following sequents

Premiss ∀x (f (f (x)) = x)
Conc ∀x∃y (x = f (y))

258
Q

Describe in text how to convert a truth table to

  • DNF
  • CNF
A
  • For DNF look at the rows where the expression becomes True, think of it as “the expression can be true at row x OR row 2 OR …..
  • For DNF look at the rows where the expression becomes False, think of it as “the expression cannot be false at row x and row 2 and …
267
Q

2016 - 4a

Let P , S and M be unary predicates and R a binary predicate. Decide for each of the sequents below whether they are valid or not, i.e., give a
proof in natural deduction or a counter-model.

Premiss ∃x (P (x) ∧ ¬M (x))

Premiss ∃y (M (y) ∧ ¬S(y))

Conc ∃z(P (z) ∧ ¬S(z))